Time Limit: 1000MS Memory Limit: 65536K

Special Judge


The City has a number of municipal buildings and a number of fallout shelters that were build specially to hide municipal workers in case of a nuclear war. Each fallout shelter has a limited capacity in terms of a number of people it can accommodate, and there’s almost no excess capacity in The City’s fallout shelters. Ideally, all workers from a given municipal building shall run to the nearest fallout shelter. However, this will lead to overcrowding of some fallout shelters, while others will be half-empty at the same time.

To address this problem, The City Council has developed a special evacuation plan. Instead of assigning every worker to a fallout shelter individually (which will be a huge amount of information to keep), they allocated fallout shelters to municipal buildings, listing the number of workers from every building that shall use a given fallout shelter, and left the task of individual assignments to the buildings’ management. The plan takes into account a number of workers in every building - all of them are assigned to fallout shelters, and a limited capacity of each fallout shelter - every fallout shelter is assigned to no more workers then it can accommodate, though some fallout shelters may be not used completely.

The City Council claims that their evacuation plan is optimal, in the sense that it minimizes the total time to reach fallout shelters for all workers in The City, which is the sum for all workers of the time to go from the worker’s municipal building to the fallout shelter assigned to this worker.

The City Mayor, well known for his constant confrontation with The City Council, does not buy their claim and hires you as an independent consultant to verify the evacuation plan. Your task is to either ensure that the evacuation plan is indeed optimal, or to prove otherwise by presenting another evacuation plan with the smaller total time to reach fallout shelters, thus clearly exposing The City Council’s incompetence.

During initial requirements gathering phase of your project, you have found that The City is represented by a rectangular grid. The location of municipal buildings and fallout shelters is specified by two integer numbers and the time to go between municipal building at the location (Xi, Yi) and the fallout shelter at the location (Pj, Qj) is Di,j = |Xi - Pj| + |Yi - Qj| + 1 minutes.


The input consists of The City description and the evacuation plan description. The first line of the input file consists of two numbers N and M separated by a space. N (1 ≤ N ≤ 100) is a number of municipal buildings in The City (all municipal buildings are numbered from 1 to N). M (1 ≤ M ≤ 100) is a number of fallout shelters in The City (all fallout shelters are numbered from 1 to M).

The following N lines describe municipal buildings. Each line contains there integer numbers Xi, Yi, and Bi separated by spaces, where Xi, Yi (-1000 ≤ Xi, Yi ≤ 1000) are the coordinates of the building, and Bi (1 ≤ Bi ≤ 1000) is the number of workers in this building.

The description of municipal buildings is followed by M lines that describe fallout shelters. Each line contains three integer numbers Pj, Qj, and Cj separated by spaces, where Pi, Qi (-1000 ≤ Pj, Qj ≤ 1000) are the coordinates of the fallout shelter, and Cj (1 ≤ Cj ≤ 1000) is the capacity of this shelter.

The description of The City Council’s evacuation plan follows on the next N lines. Each line represents an evacuation plan for a single building (in the order they are given in The City description). The evacuation plan of ith municipal building consists of M integer numbers Ei,j separated by spaces. Ei,j (0 ≤ Ei,j ≤ 1000) is a number of workers that shall evacuate from the ith municipal building to the jth fallout shelter.

The plan in the input file is guaranteed to be valid. Namely, it calls for an evacuation of the exact number of workers that are actually working in any given municipal building according to The City description and does not exceed the capacity of any given fallout shelter.


If The City Council’s plan is optimal, then write to the output the single word OPTIMAL. Otherwise, write the word SUBOPTIMAL on the first line, followed by N lines that describe your plan in the same format as in the input file. Your plan need not be optimal itself, but must be valid and better than The City Council’s one.

Sample Input

3 4

-3 3 5

-2 -2 6

2 2 5

-1 1 3

1 1 4

-2 -2 7

0 -1 3

3 1 1 0

0 0 6 0

0 3 0 2

Sample Output


3 0 1 1

0 0 6 0

0 4 0 1


Northeastern Europe 2002






#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <queue>
#include <algorithm>

using namespace std;

const int INF = 0x3f3f3f3f;

const int Max = 210;

typedef struct node
    int x,y,num;

int Map[Max][Max],Cost[Max][Max],Num[Max];

Point  Z[Max],B[Max];

int dis[Max],pre[Max],Du[Max];

bool vis[Max];

int n,m,s,t;

int ok(Point a,Point b)
    return abs(a.x-b.x)+abs(a.y-b.y)+1;

int SPFA() //判断是不是有负圈
    for(int i=0;i<=t;i++)
        dis[i] = INF;

        pre[i] = -1;

        Du[i] = 0;


    dis[t] = 0,vis[t] = true;


    Du[t] = 1;

        int u = Q.front();


        for(int i=0;i<=t;i++)
                dis[i] = dis[u]+Cost[u][i];

                pre[i] = u;




                        return i;


    return -1;

int main()

    int num;

    while(~scanf("%d %d",&n,&m))
        s= 0, t =n+m+1;




        for(int i=1;i<=n;i++) scanf("%d %d %d",&Z[i].x,&Z[i].y,&Z[i].num);

        for(int i=1;i<=m;i++) scanf("%d %d %d",&B[i].x,&B[i].y,&B[i].num);

        for(int i=1;i<=n;i++)//市政与避难所之间建图
            for(int j=1;j<=m;j++)
                Cost[i][j+n] = ok(Z[i],B[j]);

                Cost[j+n][i] = -Cost[i][j+n];

                Map[i][j+n] = Z[i].num;
        int ans = 0;

        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)

                Map[i][j+n]-= num;

                Map[j+n][i] = num;


        for(int i=1;i<=m;i++)
            Map[i+n][t] = B[i].num-Num[i];

            Map[t][i+n] = Num[i];

        ans = SPFA();



            int v = ans;


                v = pre[v];

            ans  = v;

                Map[pre[v]][v] --;


                v = pre[v];

            for(int i=1;i<=n;i++)
                for(int j=1;j<=m;j++)
                        printf(" ");


    return 0;

Evacuation Plan-POJ2175最小费用消圈算法的更多相关文章

  1. poj 2175 Evacuation Plan 最小费用流判定,消圈算法

    题目链接 题意:一个城市有n座行政楼和m座避难所,现发生核战,要求将避难所中的人员全部安置到避难所中,每个人转移的费用为两座楼之间的曼哈顿距离+1,题目给了一种方案,问是否为最优方案,即是否全部的人员 ...

  2. POJ 2157 Evacuation Plan [最小费用最大流][消圈算法]

    ---恢复内容开始--- 题意略. 这题在poj直接求最小费用会超时,但是题意也没说要求最优解. 根据线圈定理,如果一个跑完最费用流的残余网络中存在负权环,那么顺着这个负权环跑流量为1那么会得到更小的 ...

  3. POJ2175 Evacuation Plan

    Evacuation Plan Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4617   Accepted: 1218   ...

  4. nyoj 712 探 寻 宝 藏--最小费用最大流

    问题 D: 探 寻 宝 藏 时间限制: 1 Sec  内存限制: 128 MB 题目描述 传说HMH大沙漠中有一个M*N迷宫,里面藏有许多宝物.某天,Dr.Kong找到了迷宫的地图,他发现迷宫内处处有 ...

  5. poj 2195 二分图带权匹配+最小费用最大流

    题意:有一个矩阵,某些格有人,某些格有房子,每个人可以上下左右移动,问给每个人进一个房子,所有人需要走的距离之和最小是多少. 貌似以前见过很多这样类似的题,都不会,现在知道是用KM算法做了 KM算法目 ...

  6. hdu 2686 Matrix 最小费用最大流

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2686 Yifenfei very like play a number game in the n*n ...

  7. hdu 1533 Going Home 最小费用最大流

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1533 On a grid map there are n little men and n house ...

  8. 【进阶——最小费用最大流】hdu 1533 Going Home (费用流)Pacific Northwest 2004

    题意: 给一个n*m的矩阵,其中由k个人和k个房子,给每个人匹配一个不同的房子,要求所有人走过的曼哈顿距离之和最短. 输入: 多组输入数据. 每组输入数据第一行是两个整型n, m,表示矩阵的长和宽. ...

  9. SGU 185.Two shortest (最小费用最大流)

    时间限制:0.25s 空间限制:4M 题意: 在n(n<=400)个点的图中,找到并输出两条不想交的最短路.不存在输出“No sulotion”: Solution: 最小费用最大流 建图与po ...


  1. 大熊君JavaScript插件化开发------(实战篇之DXJ UI ------ ItemSelector)

    一,开篇分析 Hi,大家好!大熊君又和大家见面了,还记得前两篇文章吗.主要讲述了以“jQuery的方式如何开发插件”,以及过程化设计与面向对象思想设计相结合的方式是 如何设计一个插件的,两种方式各有利 ...

  2. nyist 78 圈水池

    http://acm.nyist.net/JudgeOnline/problem.php?pid=78 圈水池 时间限制:3000 ms  |  内存限制:65535 KB 难度:4   描述 有一个 ...

  3. 关于CSS学习的第一章

    1.CSS三种书写的方式:嵌入式.外链式.行内式 嵌入式就是将CSS写入在<style></style> 外链式将外面的CSS文件通过HTML中的标记链接过来:<link ...

  4. C# 装箱和拆箱的简单理解

    一.装箱拆箱的意义 主要用途是可以向ArrayList中添加值类型的元素 二.理解 装箱的含义:理解为可以将子类对象隐式的转化为父类对象(保留自己特有的成员,和子类重写的成员) 装箱:例子为典型的装箱 ...

  5. 阿里巴巴B2B搜索学习

    1.搜索业务 主搜索:商品搜索.商家搜索.采购搜索.app搜索 行业搜索:淘货源.淘工厂.聚好货.主题市场.品牌馆等 2.优势 由于用户多,需求强烈,收益大,所以功能.场景.架构做到极致高效. 代码复 ...

  6. 【HDU】4405 Aeroplane chess

    http://acm.hdu.edu.cn/showproblem.php?pid=4405 题意:每次可以走1~6格,初始化在第0格,走到>=n的格子就结束.还有m个传送门,表示可以从X[i] ...

  7. c++学习笔记——构造函数

    构造函数定义:每个类都分别定义了它的对象被初始化的方式,类通过一个或几个特殊的成员函数来控制其对象的初始化过程,这些函数叫做构造函数. 需要注意的几点: 1:构造函数不能被声明为const的,当我们创 ...

  8. HTTP协议状态码

    如果向您的服务器发出了某项请求要求显示您网站上的某个网页(例如,当用户通过浏览器访问您的网页或在检测工具抓取该网页时),那么,您的服务器会返回 HTTP 状态代码以响应该请求. 一些常见的状态代码为: ...

  9. 218. The Skyline Problem *HARD* -- 矩形重叠

    A city's skyline is the outer contour of the silhouette formed by all the buildings in that city whe ...

  10. Code First 数据注释

    Code First 数据注释 Julie Lerman http://thedatafarm.com 通过实体框架 Code First,可以使用您自己的域类表示 EF 执行查询.更改跟踪和更新函数 ...